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April 20, 2014

April 20, 2014

Posted by **Jacob** on Wednesday, March 6, 2013 at 7:58pm.

h(x) = x3 + 3x2 + 4 on [−3, 2]

- Applied Calculus -
**Reiny**, Wednesday, March 6, 2013 at 8:45pmI will assume you meant

f(x) = x^3 + 3x^2 + 4

f ' (x) = 3x^2 + 6x = 0 for max/min

3x(x+2) = 0

x = 0 or x = -2

f(0) = 4

f(-2) = -8 + 12 + 4 = 8

end values:

f(-3) = -27 + 27 + 4 = 4

f(2) = 8 + 12 + 4 = 24

minimum value of the function is 4

maximum value is 24

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